# Can one affine point on an elliptic curve have two Jacobians coordinates?

I have theses output on curve for jacobian coordinates which I made doubling for (3,10,1) to get (17,21,20) then I made addition for all points to get this results:

     1     3    10     1
2    17    21    20
3    11    13     7
4    20    21    14
5     2     6     6
6     9     1     8
7    14    18     8
8     6    13     2
9     0    20    11
10     3    18     9
11     3    22     2
12    10    17     5
13     9    18    20
14     3     0     8
15    13     6     6
16    10     6     5
17     1    11    20
18     3     5     9
19     0    15     2
20    13    16    22
21    19    10    13
22     8    20     4
23     9     7     1
24    14    19     6
25    22    15    11
26     5     4    21
27    13    22    14


but when use scalar multiplication double-and-add Algorithm to get 4P from 2P then I got this coordinates (10,9,12) instead of (20,21,14)? I believe that two points are on the curve as 4P because I revert them to affine and I got it correct affine coordinates for both are the same which is (17,3)

My question does One affine curve point has two jacobians?

Yes, much like $$\displaystyle\frac32$$ and $$\displaystyle\frac64$$ are two different representations of the same rational number, a point on an Elliptic Curve has multiple different Jacobian representations.
$$(X_0,Y_0,Z_0)$$ and $$(X_1,Y_1,Z_1)$$ with $$Z_0\ne0$$ and $$Z_1\ne0$$ represent the same point in Jacobian coordinates if and only if $${Z_0}^2\,X_1={Z_1}^2\,X_0$$ and $${Z_0}^3\,Y_1={Z_1}^3\,Y_0$$, evaluated in the base field.
That can be rewritten $$\displaystyle\frac{X_0}{{Z_0}^2}=\frac{X_1}{{Z_1}^2}$$ and $$\displaystyle\frac{Y_0}{{Z_0}^3}=\frac{Y_1}{{Z_1}^3}$$, where the division is in the base field. The two quantities (on left and right of the previous and) are the Cartesian (aka affine) coordinates.
Curve point (other than the Elliptic Curve's neutral point/point at infinity) with Jacobian coordinates $$(x,y,1)$$ is the point with Cartesian coordinates $$(x,y)$$.